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# If an integer n is to be chosen at random from the integers

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If an integer n is to be chosen at random from the integers [#permalink]

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11 Sep 2007, 13:18
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If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4

B. 3/8

C. 1/2

D. 5/8

E. 3/4

I was able to solve this problem by brute force i.e. by listing n(n + 1)(n + 2) sets i.e. {1,2,3} {2,3,4} {3,4,5} etc. Obviously this is a dumb method of solving the question... it took a large amount of time. There has to be a 2 minute method right?
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Re: PS: Set 4, Math 9 [#permalink]

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11 Sep 2007, 18:19
gluon wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4

B. 3/8

C. 1/2

D. 5/8

E. 3/4

I was able to solve this problem by brute force i.e. by listing n(n + 1)(n + 2) sets i.e. {1,2,3} {2,3,4} {3,4,5} etc. Obviously this is a dumb method of solving the question... it took a large amount of time. There has to be a 2 minute method right?

I got 5/8 and it is a bit time consuming.

n(n + 1)(n + 2)
First, this tells you that if n is multiple of 8, then it is surely divisible by 8. Go further, if n is multiple of 4, you can also tell that this is divisible by 8 since 4*even is divisible by 8. Go even further, you can see that if n is even, then it is divisible by 8. The reason for this is that if n is eve, then
n*(n+2) will always be even*(even+2) and (even+2) is always divisible by 4 because
(even +2)/even = odd + odd = even
So all even number works.
all odd number that -1 and equal to 8 also works.
so all even = 96/2 = 48
all odd as mentioned = 96/8 = 12
Total = 48+12
Ans = 60/96 = 5/8
Director
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Re: PS: Set 4, Math 9 [#permalink]

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11 Sep 2007, 18:57
bkk145 wrote:
gluon wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

I was able to solve this problem by brute force i.e. by listing n(n + 1)(n + 2) sets i.e. {1,2,3} {2,3,4} {3,4,5} etc. Obviously this is a dumb method of solving the question... it took a large amount of time. There has to be a 2 minute method right?

I got 5/8 and it is a bit time consuming.

n(n + 1)(n + 2)
First, this tells you that if n is multiple of 8, then it is surely divisible by 8. Go further, if n is multiple of 4, you can also tell that this is divisible by 8 since 4*even is divisible by 8. Go even further, you can see that if n is even, then it is divisible by 8. The reason for this is that if n is eve, then
n*(n+2) will always be even*(even+2) and (even+2) is always divisible by 4 because

(even +2)/even = odd + odd = even
So all even number works.
all odd number that -1 and equal to 8 also works.
so all even = 96/2 = 48
all odd as mentioned = 96/8 = 12
Total = 48+12
Ans = 60/96 = 5/8

agree with 5/8. but i am confused with the highlited part.
VP
Joined: 10 Jun 2007
Posts: 1444
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Re: PS: Set 4, Math 9 [#permalink]

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11 Sep 2007, 19:06
Fistail wrote:
bkk145 wrote:
gluon wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

I was able to solve this problem by brute force i.e. by listing n(n + 1)(n + 2) sets i.e. {1,2,3} {2,3,4} {3,4,5} etc. Obviously this is a dumb method of solving the question... it took a large amount of time. There has to be a 2 minute method right?

I got 5/8 and it is a bit time consuming.

n(n + 1)(n + 2)
First, this tells you that if n is multiple of 8, then it is surely divisible by 8. Go further, if n is multiple of 4, you can also tell that this is divisible by 8 since 4*even is divisible by 8. Go even further, you can see that if n is even, then it is divisible by 8. The reason for this is that if n is eve, then
n*(n+2) will always be even*(even+2) and (even+2) is always divisible by 4 because

(even +2)/even = odd + odd = even
So all even number works.
all odd number that -1 and equal to 8 also works.
so all even = 96/2 = 48
all odd as mentioned = 96/8 = 12
Total = 48+12
Ans = 60/96 = 5/8

agree with 5/8. but i am confused with the highlited part.

For a number to be divisible by 4, it must be divisible by 2 two times. I am trying to show that (even +2) is divisible by 4 by showing that it is divisible by 2 two times.
1st time: (even+2)/2= even
2nd time: even/even = odd
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Re: PS: Set 4, Math 9 [#permalink]

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11 Sep 2007, 19:16
bkk145 wrote:
gluon wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4

B. 3/8

C. 1/2

D. 5/8

E. 3/4

I was able to solve this problem by brute force i.e. by listing n(n + 1)(n + 2) sets i.e. {1,2,3} {2,3,4} {3,4,5} etc. Obviously this is a dumb method of solving the question... it took a large amount of time. There has to be a 2 minute method right?

I got 5/8 and it is a bit time consuming.

n(n + 1)(n + 2)
First, this tells you that if n is multiple of 8, then it is surely divisible by 8. Go further, if n is multiple of 4, you can also tell that this is divisible by 8 since 4*even is divisible by 8. Go even further, you can see that if n is even, then it is divisible by 8. The reason for this is that if n is eve, then
n*(n+2) will always be even*(even+2) and (even+2) is always divisible by 4 because
(even +2)/even = odd + odd = even
So all even number works.
all odd number that -1 and equal to 8 also works.
so all even = 96/2 = 48
all odd as mentioned = 96/8 = 12
Total = 48+12
Ans = 60/96 = 5/8

Fantabulous ---(fantastic + fabulous) explanation...
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11 Sep 2007, 19:25
awsome..i had figured out that all even numbers would work..but totally forgot that 7, 15 i.e (8n-1) would also work...

48+12=60/96 5/8
Director
Joined: 03 May 2007
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Re: PS: Set 4, Math 9 [#permalink]

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11 Sep 2007, 22:39
bkk145 wrote:
Fistail wrote:
bkk145 wrote:
gluon wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

I was able to solve this problem by brute force i.e. by listing n(n + 1)(n + 2) sets i.e. {1,2,3} {2,3,4} {3,4,5} etc. Obviously this is a dumb method of solving the question... it took a large amount of time. There has to be a 2 minute method right?

I got 5/8 and it is a bit time consuming.

n(n + 1)(n + 2)
First, this tells you that if n is multiple of 8, then it is surely divisible by 8. Go further, if n is multiple of 4, you can also tell that this is divisible by 8 since 4*even is divisible by 8. Go even further, you can see that if n is even, then it is divisible by 8. The reason for this is that if n is eve, then
n*(n+2) will always be even*(even+2) and (even+2) is always divisible by 4 because

(even +2)/even = odd + odd = even
So all even number works.
all odd number that -1 and equal to 8 also works.
so all even = 96/2 = 48
all odd as mentioned = 96/8 = 12
Total = 48+12
Ans = 60/96 = 5/8

agree with 5/8. but i am confused with the highlited part.

For a number to be divisible by 4, it must be divisible by 2 two times. I am trying to show that (even +2) is divisible by 4 by showing that it is divisible by 2 two times.
1st time: (even+2)/2= even
2nd time: even/even = odd

oh no. what happend to me? i saw the red part as even + odd = even.

now it is perfect. how come? either you did some magic or i was in second life!!

agree that it was/is "Fantabulous"
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11 Sep 2007, 22:58
Between 1 and 96 --> 96 numbers
# of even numbers = 48
# of odd numbers = 48

To be divisible by 8, the number must be divisible by 2 three times.

Possibilites for n = every even integers --> 48
We don't need to consider odd for n+1 as that is equal to n being even.
We also don't need to consider even for n+2 as that is equal to n being even.
But we do need to consider multiples of 8 for n+1. # of multiple of 8 = (95-7)/8 + 1 = 12

So total # of winning cases = 48+12 = 60

P = 60/96 = 5/8
11 Sep 2007, 22:58
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